The definition of the metric tensor is the matrix composed of zero modes’ overlaps. At large separations between dyons, the four approximately zero modes per dyon are $F_{\mu \alpha }^{(i)}$, $\alpha =1,2,3,4$, where $F^{(i)}$ is the field strength of the $i$th dyon whose scale parameter $\nu$ has to be adjusted by the Coulomb field of other dyons. This is why the first correction to the metric is always Coulomb-like. It is remarkable that for different-kind dyons there are no further corrections.

In principle, one can think of a quantum anomaly leading to an uncomplete cancellation of boson and ghost determinants, as happens in some supersymmetric examples [36]. We do not see a reason for the anomaly in our case, however. Therefore, unless proven otherwise, we shall assume that the cancellation is exact.

Temperature dependence of the string tension appears when one adds, e.g., the perturbative potential energy into consideration. Since its relative contribution is $\sim T^4$, one expects that it modifies seriously Eq. (84) only close to the phase transition temperature where the electric string tension vanishes.

In our linearized approach, there are combinatorial contributions $\sim \exp (-|{\mathbf{z}}_1-{\mathbf{z}}_2|M(\sqrt{{\mathcal{M}}^{(l_1)}}+\sqrt{{\mathcal{M}}^{(l_2)}}+\dots ))$ to the correlator, stemming from expanding the exponent in Eq. (79) and its analogs for arbitrary $k$ in higher powers of $\exp (-|{\mathbf{z}}_1-{\mathbf{z}}_2|M\sqrt{{\mathcal{M}}^{(l)}})$. Starting from $k\ge 3$, there is a danger that a combination of lower eigenvalues may give a larger contribution to the correlator than a single contribution of ${\mathcal{M}}^{(k)}$, for example, $2\sqrt{{\mathcal{M}}^{(1)}}<\sqrt{{\mathcal{M}}^{(3)}}$. For this particular case, we have proved that the square of the ${\mathcal{M}}^{(1)}$ contribution decouples from the $k=3$ correlator for all $N$. However, we have not proved it for all $k$ and all possible combinations of eigenvalues whose sum is less than $\sqrt{{\mathcal{M}}^{(k)}}$: Probably, certain generalizations of the orthogonality relations (89) are responsible for it. A more appropriate approach would be to solve the full nonlinear equations. We note that combinatorial contributions leading to larger string tensions are generally present and can be considered as string excitations.